已知x,y∈R,求证:x^2+y^2≥xy+x+y-1
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/21 00:22:54
已知x,y∈R,求证:x^2+y^2≥xy+x+y-1
![已知x,y∈R,求证:x^2+y^2≥xy+x+y-1](/uploads/image/z/16959257-17-7.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%E2%88%88R%2C%E6%B1%82%E8%AF%81%3Ax%5E2%2By%5E2%E2%89%A5xy%2Bx%2By-1)
(x2+y2)-(xy+x+y-1)
=(1/2)*[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]
=(1/2)*[(x-y)^2+(x-1)^2+(y-1)^2]
因为(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0
(三项都取=号,有解x=y=1)
所以
(x2+y2)-(xy+x+y-1)≥0
x^2+y^2≥xy+x+y-1
=(1/2)*[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]
=(1/2)*[(x-y)^2+(x-1)^2+(y-1)^2]
因为(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0
(三项都取=号,有解x=y=1)
所以
(x2+y2)-(xy+x+y-1)≥0
x^2+y^2≥xy+x+y-1
已知x,y∈R+,且x+y=1,求证:xy+1xy≥174
已知x.y∈R,求证x2+y2+1≥x+y+xy
已知x,y属于正R,且x+2y=1,求证xy=
已知x,y∈R*,x+y=1,求证2/x+1/y≥3+2根号2
已知x,y∈R*,x+y=xy,求u=x+2y最小值
已知x>y>0,xy=1,求证(x^2+y^2)/(x-y)≥2根号2
已知x,y是正实数,且xy-x-y=1,求证x+y≥2+√2
已知x.y属于R,用向量法证明x*x+y*y>=2xy
已知xy为任意实数,求证x^4+y^4≥1/2xy(x+y)^2
x,y属于R*,且x+y=1,求证:(1)(x+1/x)(y+1/y)≥25/4 (2)(x+1/x)^2+(y+1/y
已知xy都是正实数,且X+Y>2,求证1+X/Y
已知x,y均为正实数.(1)求证:2xy/x+y